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Lecture 2: Trigonometric Functions: Identities 2.1 Some basic identities For any angle θ, (cos(θ), sin(θ)) is a point on the unit circle x2 + y 2 = 1. Hence we must have sin2 (θ) + cos2 (θ) = 1 for all angles θ. Example If sin(θ) = 14 , then cos2 (θ) = 1 − sin2 (θ) = 1 − Thus 1 15 = . 16 16 √ 15 4 or √ 15 cos(θ) = − . 4 Since sin(θ) > 0, θ must lie in either the first or second quadrant. If in the first, then √ 15 ; cos(θ) = 4 cos(θ) = if in the second, then √ cos(θ) = − 15 . 4 Dividing both sides of sin2 (θ) + cos2 (θ) = 1 by cos2 (θ) gives us tan2 (θ) + 1 = sec2 (θ) for all θ. The geometry of the circle shows us that, for any angle θ, the point (cos(−θ), sin(−θ)) is the point (cos(θ), sin(θ)) reflected over the x-axis, an action which does not change the value of the x-coordinate, but does change the sign of the y-coordinate. That is, cos(−θ) = cos(θ) and sin(−θ) = − sin(θ) for all θ. 2-1 Lecture 2: Trigonometric Functions: Identities Example We have π π √3 cos − = cos = 6 6 2 and π π 1 sin − = − sin =− . 6 6 2 Since the unit circle has circumference 2π, we have sin(θ + 2nπ) = sin(θ) and cos(θ + 2nπ) = cos(θ) for any integer n. We say that the functions sin(θ) and cos(θ) have period 2π. Example We have sin and cos 8π 3 8π 3 = sin 8π − 2π 3 = cos 8π − 2π 3 √ = sin 2π 3 = = cos 2π 3 1 =− . 2 3 2 2.2 Addition and subtraction formulas For any real numbers x and y, cos(x + y) = cos(x) cos(y) − sin(x) sin(y) and sin(x + y) = sin(x) cos(y) + cos(x) sin(y). These are the addition formulas for sine and cosine. Example Since π π 3π 4π 7π + = + = , 4 3 12 12 12 it follows that cos 7π 12 = cos π 4 cos π 3 − sin π 4 sin π 3 √ √ 1 3 1− 3 1 1 √ −√ = =√ 2 2 2 2 22 2-2 Lecture 2: Trigonometric Functions: Identities 2-3 and sin 7π 12 = sin π 4 cos π 3 + cos π 4 sin π 3 √ √ 1+ 3 1 1 1 3 √ . = =√ +√ 22 2 2 2 2 It follows from the addition formulas that for any real numbers x and y, cos(x − y) = cos(x) cos(y) + sin(x) sin(y) and sin(x − y) = sin(x) cos(y) − cos(x) sin(y). These are the subtraction formulas for sine and cosine. Example Since 3π 4π π π π − = − =− , 4 3 12 12 12 it follows that √ √ π π π π π 1 1 1 3 1+ 3 √ = cos cos + sin sin =√ +√ = cos − 12 4 3 4 3 22 2 2 2 2 and √ √ π π π π π 1 1 1 3 1− 3 √ . sin − = sin cos − cos sin =√ −√ = 12 4 3 4 3 22 2 2 2 2 2.3 Double and half angle formulas From the addition formula for cosine we have, for any real number x, cos(2x) = cos(x + x) = cos(x) cos(x) − sin(x) sin(x) = cos2 (x) − sin2 (x) and from the addition formula for sine we have, for any real number x, sin(2x) = sin(x + x) = sin(x) cos(x) + cos(x) sin(x) = 2 sin(x) cos(x), the double angle formulas for sine and cosine. Now sin2 (x) = 1 − cos2 (x), so from the double angle formula for cosine we have cos(2x) = cos2 (x) − (1 − cos2 (x)) = 2 cos2 (x) − 1. Solving for cos(x), we have cos2 (x) = 1 + cos(2x) , 2 Lecture 2: Trigonometric Functions: Identities 2-4 the half-angle formula for cosine. Similarly, using cos2 (x) = 1 − sin2 (x), we have cos(2x) = (1 − sin2 (x)) − sin2 (x) = 1 − 2 sin2 (x), from which we obtain sin2 (x) = 1 − cos(2x) , 2 the half-angle formula for sine. Example We have π π 1 + cos 2 6 cos = 12 2 and so π cos = 12 √ 3 √ 2 + 3 2 = , 2 4 1+ = p 2+ 2 √ Similarly, π π 1 − cos 6 = sin2 12 2 3 . √ 3 √ 2 = 2 − 3, 2 4 1− = and so p √ π 2− 3 = . sin 12 2 Note that we used the positive square roots because we know that both cos π sin 12 are positive. π 12 and 2.4 Shift formulas We may also derive a number of shift formulas from the addition and subtractions formulas. For example, π π π cos x − = cos(x) cos + sin(x) sin = sin(x) 2 2 2 for all x and π π π sin x − = sin(x) cos − cos(x) sin = − cos(x) 2 2 2 for all x. Also, cos(x + π) = cos(x) cos(π) − sin(x) sin(π) = − cos(x) for all x and sin(x + π) = sin(x) cos(π) + cos(x) sin(π) = − sin(x) for all x. As a consequence, note that − sin(x) sin(x + π) tan(x + π) = = = tan(x) cos(x + π) − cos(x) for all x. That is, tan(x) has period π.